Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $t = \dfrac{q^2 + 14q + 45}{5q - 15} \div \dfrac{q^2 + 9q}{4q - 12} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{q^2 + 14q + 45}{5q - 15} \times \dfrac{4q - 12}{q^2 + 9q} $ First factor the quadratic. $t = \dfrac{(q + 9)(q + 5)}{5q - 15} \times \dfrac{4q - 12}{q^2 + 9q} $ Then factor out any other terms. $t = \dfrac{(q + 9)(q + 5)}{5(q - 3)} \times \dfrac{4(q - 3)}{q(q + 9)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (q + 9)(q + 5) \times 4(q - 3) } { 5(q - 3) \times q(q + 9) } $ $t = \dfrac{ 4(q + 9)(q + 5)(q - 3)}{ 5q(q - 3)(q + 9)} $ Notice that $(q - 3)$ and $(q + 9)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ 4\cancel{(q + 9)}(q + 5)(q - 3)}{ 5q(q - 3)\cancel{(q + 9)}} $ We are dividing by $q + 9$ , so $q + 9 \neq 0$ Therefore, $q \neq -9$ $t = \dfrac{ 4\cancel{(q + 9)}(q + 5)\cancel{(q - 3)}}{ 5q\cancel{(q - 3)}\cancel{(q + 9)}} $ We are dividing by $q - 3$ , so $q - 3 \neq 0$ Therefore, $q \neq 3$ $t = \dfrac{4(q + 5)}{5q} ; \space q \neq -9 ; \space q \neq 3 $